Problem 12 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C} y d x+x d y\) on the given curve \(C\) between \((0,0)\) and \((1,1)\). $$ y=x $$

Step-by-Step Solution

Answer

The value of the integral is 1.

Step 1: Identify the Curve Equation

The curve given is \( y = x \). This means that the path \( C \) is a straight line from point \( (0,0) \) to \( (1,1) \) along the line \( y = x \).

Step 2: Parametrize the Line

Since \( y = x \), we can parametrize the curve using the parameter \( t \) where \( x = t \) and \( y = t \). Hence, as \( t \) varies from 0 to 1, both \( x \) and \( y \) vary, tracing the path from \( (0,0) \) to \( (1,1) \).

Step 3: Determine Expressions for dx and dy

Since \( x = t \) and \( y = t \), then \( dx = dt \) and \( dy = dt \).

Step 4: Substitute into the Integral

Substitute \( x = t \), \( y = t \), \( dx = dt \), and \( dy = dt \) into the integral: \[ \int_C y \, dx + x \, dy = \int_0^1 t \, dt + t \, dt.\]

Step 5: Simplify the Integral

Combine the terms in the integral: \[ \int_0^1 t \, dt + t \, dt = \int_0^1 (2t) \, dt. \]

Step 6: Evaluate the Integral

Calculate the integral: \[ \int_0^1 (2t) \, dt = 2 \left[ \frac{t^2}{2} \right]_0^1 = 2 \left( \frac{1^2}{2} - \frac{0^2}{2} \right) = 2 \cdot \frac{1}{2} = 1. \]

Key Concepts

Parameterization

Parameterization is the process of describing a curve by expressing the coordinates of its points as functions of a single variable, known as the parameter. In the context of line integrals, parameterization is essential to simplify the integration process. By representing both x and y in terms of a parameter t, we transform a curve into a simpler, often one-dimensional path that is easier to integrate over.

In this exercise, we start with the equation of a line: \( y = x \). We choose a parameter \( t \) that captures the points along the line from \( (0,0) \) to \( (1,1) \).

• Let \( x = t \), so \( y = t \) directly follows from \( y = x \), covering the entire line as \( t \) ranges from 0 to 1.

- With the parameterization, every point (x, y) along the curve is now expressed as (t, t), transforming a 2D path into a straightforward 1D path.

Curve Integration

Curve integration, also known as line integration, involves integrating a function over a curve.

In this problem, we integrate over the path \( C \) from point \( (0,0) \) to \( (1,1) \) along the line \( y = x \). Once we have parameterized our curve, we need to determine differentials based on the parameter:

• Since \( x = t \) and \( y = t \), it follows that \( dx = dt \) and \( dy = dt \).

- This reduction simplifies our integral as it translates the problem of integrating over a complex path to one of integrating along a simpler parameter, \( t \).

This makes evaluating complex curves manageable and less error-prone. All interactions of\( x \) and \( y \) become interactions of \( t \), allowing the integral to focus on changes in \( t \) alone.

Integral Evaluation

Integral evaluation involves finding the numerical value of the integral over the parameterized curve. In this exercise, it involves completing the calculation after substitution of the parameters.

After parameterizing the curve using \( t \), we replace \( x \), \( y \), \( dx \), and \( dy \) in the line integral \( \int_{C} y \, dx + x \, dy \) with \( t \), \( t \), \( dt \) and \( dt \), respectively. This results in:

• \( \int_0^1 t \, dt + t \, dt = \int_0^1 (2t) \, dt \).

By simplifying, we combine the terms and then calculate the definite integral of \( 2t \) from 0 to 1.
- The integral evaluates to: \( 2 \left[ \frac{t^2}{2} \right]_0^1 = 2 \cdot \frac{1}{2} = 1 \).

Thus, evaluating line integrals through parameterization and transforming them to simple integrals of \( t \) ensures efficiency and clarity in solving such calculus problems.